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Archy [21]
3 years ago
10

An inventor claims to have devised a closed cyclic engine which exchanges heat with cold and hot reservoirs at 25 and 350 ◦C, re

spectively, and produces 0.45 kJ of work for every kJ of heat extracted from the hot reservoir. Is this claim believable?
Chemistry
1 answer:
kati45 [8]3 years ago
7 0

Explanation:

The given data is as follows.

T_{c} = 25^{o}C,  T_{h} = 350^{o}C

Work produced per kJ of heat extracted from hot reservoir = 0.45 kJ = Efficiency

If the device is Carnot cycle then its efficiency will be maximum and its value will be equal to [1 - (\frac{T_{c}}{T_{h}} )]

Using this relation we will calculate the efficiency as follows.

                 Efficiency = [1 - (\frac{T_{c}}{T_{h}} )]

                        = 1 - (\frac{25}{350})

                       = 0.928

Hence, it means that this type of device is possible and the claim is also believable.

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Which answer best defines solute ?
Bad White [126]

Answer:

a substance that dissolves in another substance

Explanation:

a substance into which another substance dissolves

4 0
3 years ago
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The molecular weight of table salt, NaCl, is 58.5 g/mol. A tablespoon of salt weighs 6.37 grams. Calculate the number of moles o
cestrela7 [59]

Answer:

The number of moles of salt in one tablespoon is = <u>0.11 mole</u>

<u>Grams </u>cancel each other.

Explanation:

<u>Moles</u> : It is the unit of quantity . It is the mass of the substance present in exactly 12g of C-12.

moles=\frac{given\ mass}{Molar\ mass}

<u>Moles Calculation:</u>

Given mass = 6.37  gram

Molar mass = 58.5 g/mol

moles=\frac{6.37}{58.5}

= 0.1088

= 0.11 mole

<u>Units calculation</u>

moles=\frac{given\ mass}{Molar\ mass}

moles=\frac{g}{g/mole}

moles=\frac{g}{g}\times mole

<u>g ang g cancels each other </u>

moles = moles

<u>Hence unit = gram (g ) cancel each other.</u>

3 0
3 years ago
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The triple point of nitrogen occurs at a temperature of 63.1 K and a pressure of 0.127 atm. Its normal boiling point is 77.4 K.
yulyashka [42]

Nitrogen has a normal boiling point of 77.4 K and a melting point (at 1 atm) of 63.2 K. Its critical temperature is 126.2 K, and its critical pressure is 2.55 * 104 torr. It has a triple point at 63.1 K and 94.0 torr.

<h3>What is the triple point?</h3>

The temperature and pressure at which the solid, liquid, and vapour phases of a pure substance can coexist in equilibrium.

- Normal melting point: 63.2 K.

- Normal boiling point: 77.4 K.

- Triple point: 0.127 atm and 63.1 K.

- Critical point: 33.5 atm and 126.0 K.

In such a way:

- N2 does not exist as a liquid at pressures below 0.127 atm: that is because below this point, solid N2 exists only (triple point).

- N2 is a solid at 16.7 atm and 56.5 K: that is because it is above the triple point, below the critical point and below the normal melting point.

- N2 is a liquid at 1.00 atm and 73.9 K: that is because it is above the triple point, below the critical point and below the normal boiling point.

- N2 is a gas at 0.127 atm and 84.0 K: that is because it is above the triple point temperature at the triple point pressure.

Learn more about triple point here:

brainly.com/question/23307017

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5 0
2 years ago
9. Carbon-14 has a half-life of 5,730 years. How many years have passed after 3<br> half-lives?
ella [17]

Answer:

120 gram sample of a radioactive element, how many grams of that element will be left after 3 half-lives have passed? If you have a 300

Explanation:

Hope this helps!

7 0
3 years ago
A chemical compound has a molecular weight of 89.05 g/mole. 1.400 grams of this compound underwent complete combustion under con
Nataly_w [17]

Answer:

\Delta _{comb}H=-2,265\frac{kJ}{mol}

Explanation:

Hello!

In this case, for such calorimetry problem, we can notice that the combustion of the compound releases the heat which causes the increase of the temperature by 11.95 °C, it means that we can write:

Q _{comb}=-C_{calorimeter}\Delta T_{calorimeter}

In such a way, we can compute the total released heat due to the combustion considering the calorimeter specific heat and the temperature raise:

Q _{comb}=-2980\frac{J}{\°C} *11.95\°C\\\\Q _{comb}=-35,611J

Next, we compute the molar heat of combustion of the compound by dividing by the moles, considering 1.400 g were combusted:

n=1.400g*\frac{1mol}{89.05g} =0.01572mol

Thus, we obtain:

\Delta _{comb}H=\frac{Q_{comb}}{n}=\frac{-35,611J}{0.01572mol}  \\\\\Delta _{comb}H=-2,265,331\frac{J}{mol}*\frac{1kJ}{1000J}  \\\\\Delta _{comb}H=-2,265\frac{kJ}{mol}

Best regards!

7 0
3 years ago
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